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Equation 1 · From Origins to Frontier: A History of AI Memory Systems and the Bandwidth Wall

What does this equation mean?

compute(t)bandwidth(t)  =  compute(0)bandwidth(0)⋅(3.01.6)t/2,\frac{\text{compute}(t)}{\text{bandwidth}(t)} \;=\; \frac{\text{compute}(0)}{\text{bandwidth}(0)} \cdot \left(\frac{3.0}{1.6}\right)^{t/2},

Read the formula alongside the article passage below. Each part has a deeper page with its role in the equation, the supporting passage and nearby citations.

Start withcompute(0)
Divide bybandwidth(0)
This relates tofraccompute(t)bandwidth(t)
How to read the two sides of this formula. Follow the article passage for the meaning of each quantity.

This equation states an equality: the expressions on both sides have the same value under the article’s assumptions. Read the equation part by part below; each part has a contextual explanation and a link to its mathematical background.

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tt

Symbol t

t is an argument of the function-like quantity on the left; its role is set by that function’s stated inputs.

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=

=

The expressions on both sides represent the same quantity under the stated assumptions.

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fraction

fraction

Divide the expression above the line by the one below it.

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multiplication

multiplication

Multiply the quantities on either side.

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superscript

superscript

A raised number can be a power. When it is a label or bound, it selects a case or the upper limit of a sum; the formula’s structure distinguishes these uses.

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compute(t)\text{compute}(t)

Numerator: compute(t)

The complete quantity above the fraction bar.

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bandwidth(t)\text{bandwidth}(t)

Denominator: bandwidth(t)

The complete quantity below the fraction bar; it must be nonzero for this division.

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compute(0)\text{compute}(0)

Numerator: compute(0)

The complete quantity above the fraction bar.

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bandwidth(0)\text{bandwidth}(0)

Denominator: bandwidth(0)

The complete quantity below the fraction bar; it must be nonzero for this division.

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3.03.0

Numerator: 3.0

The complete quantity above the fraction bar.

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1.61.6

Denominator: 1.6

The complete quantity below the fraction bar; it must be nonzero for this division.

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How to interpret it

With a fixed numerator, increasing a nonzero denominator reduces the fraction. Read it with the definitions, units, and assumptions supplied by the article.

What the article says around this equation

The scale of that abstract argument, made concrete decades later, is stark. Gholami and colleagues, surveying twenty years of server hardware, report that peak hardware FLOPS scaled at roughly 3.0 times every two years while DRAM bandwidth scaled at only 1.6 times and interconnect bandwidth at 1.4 times over the same interval [ 13 ] . Those are compounding rates, and compounding rates diverge violently over long periods. If the two multipliers applied uniformly across a full twenty years — an extrapolation the source itself does not perform, offered here only to size the shape of the problem, not as a number the cited paper states — the ratio between available arithmetic and available…
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The scale of that abstract argument, made concrete decades later, is stark. Gholami and colleagues, surveying twenty years of server hardware, report that peak hardware FLOPS scaled at roughly 3.0 times every two years while DRAM bandwidth scaled at only 1.6 times and interconnect bandwidth at 1.4 times over the same interval [ 13 ] . Those are compounding rates, and compounding rates diverge violently over long periods. If the two multipliers applied uniformly across a full twenty years — an extrapolation the source itself does not perform, offered here only to size the shape of the problem, not as a number the cited paper states — the ratio between available arithmetic and available bandwidth grows as compute(t)bandwidth(t)  =  compute(0)bandwidth(0)⋅(3.01.6)t/2\frac{\text{compute}(t)}{\text{bandwidth}(t)} \;=\; \frac{\text{compute}(0)}{\text{bandwidth}(0)} \cdot \left(\frac{3.0}{1.6}\right)^{t/2}. with t measured in years. At t = 20 that factor works out to (3.0/1.6)^{10} ≈\approx 540 : a roughly five-hundredfold widening of the gap in raw compounding terms, and that is before accounting for the interconnect term separately. No single memory technology closes a gap shaped like that on its own. Every technology discussed below narrows it at one particular tier, for one particular class of workload, for a while — which is exactly why the industry kept needing a new one.

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Sources cited in the surrounding passage

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